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Equivalent ratios in tables

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5541926
A smoothie recipe keeps the same ratio of fruit to yogurt. <table> <tr><th>Fruit (cups)</th><th>Yogurt (cups)</th></tr> <tr><td>2</td><td>3</td></tr> <tr><td>4</td><td>?</td></tr> </table> What number belongs in the missing table entry?

Hints

- Compare the two fruit entries first. - Equivalent-ratio rows scale both quantities by the same factor. - Apply that same change to the yogurt entry.

Solution

1. The fruit amount changes from \(2\) cups to \(4\) cups, which is a factor of \(2\). 2. Multiply the yogurt amount by the same factor: \(3 \times 2 = 6\).

Answer

\(6\) cups
5162886
Three robots move at constant rates. Complete the table. <table> <tr> <td>Robot</td> <td>Distance in \(1\,\text{min}\)</td> <td>Distance in \(2\,\text{min}\)</td> <td>Distance in \(5\,\text{min}\)</td> </tr> <tr> <td>Racer</td> <td>\(50\,\text{ft}\)</td> <td></td> <td></td> </tr> <tr> <td>Walker</td> <td>\(10\,\text{ft}\)</td> <td></td> <td></td> </tr> <tr> <td>Crawler</td> <td>\(1\,\text{ft}\)</td> <td></td> <td></td> </tr> </table>

Hints

- Work across one row at a time. - Double the one-minute distance to find the two-minute distance. - Multiply the one-minute distance by \(5\) to find the five-minute distance.

Solution

1. Double each one-minute distance to find the two-minute distances. 2. Multiply each one-minute distance by \(5\) to find the five-minute distances. 3. The completed table is: <table> <tr><td>Robot</td><td>Distance in \(1\,\text{min}\)</td><td>Distance in \(2\,\text{min}\)</td><td>Distance in \(5\,\text{min}\)</td></tr> <tr><td>Racer</td><td>\(50\,\text{ft}\)</td><td>\(100\,\text{ft}\)</td><td>\(250\,\text{ft}\)</td></tr> <tr><td>Walker</td><td>\(10\,\text{ft}\)</td><td>\(20\,\text{ft}\)</td><td>\(50\,\text{ft}\)</td></tr> <tr><td>Crawler</td><td>\(1\,\text{ft}\)</td><td>\(2\,\text{ft}\)</td><td>\(5\,\text{ft}\)</td></tr> </table>

Answer

<table> <tr><td>Robot</td><td>Distance in \(1\,\text{min}\)</td><td>Distance in \(2\,\text{min}\)</td><td>Distance in \(5\,\text{min}\)</td></tr> <tr><td>Racer</td><td>\(50\,\text{ft}\)</td><td>\(100\,\text{ft}\)</td><td>\(250\,\text{ft}\)</td></tr> <tr><td>Walker</td><td>\(10\,\text{ft}\)</td><td>\(20\,\text{ft}\)</td><td>\(50\,\text{ft}\)</td></tr> <tr><td>Crawler</td><td>\(1\,\text{ft}\)</td><td>\(2\,\text{ft}\)</td><td>\(5\,\text{ft}\)</td></tr> </table>
5168116
Complete the table for movie tickets. <table> <tr> <td>\(1\) movie ticket</td> <td>\(\$9.00\)</td> </tr> <tr> <td>\(3\) movie tickets</td> <td>?</td> </tr> <tr> <td>\(7\) movie tickets</td> <td>?</td> </tr> </table>

Hints

- What is the price of one ticket? - Multiply the one-ticket price by the number of tickets. - Which operation represents equal prices repeated several times?

Solution

1. Find the cost of \(3\) tickets: \(3 \times \$9.00 = \$27.00\). 2. Find the cost of \(7\) tickets: \(7 \times \$9.00 = \$63.00\). 3. Complete the table. <table> <tr><td>\(1\) movie ticket</td><td>\(\$9.00\)</td></tr> <tr><td>\(3\) movie tickets</td><td>\(\$27.00\)</td></tr> <tr><td>\(7\) movie tickets</td><td>\(\$63.00\)</td></tr> </table>

Answer

<table> <tr><td>\(1\) movie ticket</td><td>\(\$9.00\)</td></tr> <tr><td>\(3\) movie tickets</td><td>\(\$27.00\)</td></tr> <tr><td>\(7\) movie tickets</td><td>\(\$63.00\)</td></tr> </table>
5204256
At a garden store, \(5\) seed packets cost \(\$4\). Find the cost of \(10\), \(15\), and \(25\) seed packets. Show the equivalent ratios in a table.

Hints

- Determine how many groups of \(5\) are in each requested quantity. - Multiply the cost of one group by the same factor. - Keep the packet and cost values aligned in the table.

Solution

1. Ten packets are \(2\) groups of \(5\), so they cost \(2 \times \$4 = \$8\). 2. Fifteen packets are \(3\) groups of \(5\), so they cost \(3 \times \$4 = \$12\). 3. Twenty-five packets are \(5\) groups of \(5\), so they cost \(5 \times \$4 = \$20\). <table> <tr><th>Seed packets</th><th>Cost</th></tr> <tr><td>\(5\)</td><td>\(\$4\)</td></tr> <tr><td>\(10\)</td><td>\(\$8\)</td></tr> <tr><td>\(15\)</td><td>\(\$12\)</td></tr> <tr><td>\(25\)</td><td>\(\$20\)</td></tr> </table>

Answer

<table> <tr><th>Seed packets</th><th>Cost</th></tr> <tr><td>\(5\)</td><td>\(\$4\)</td></tr> <tr><td>\(10\)</td><td>\(\$8\)</td></tr> <tr><td>\(15\)</td><td>\(\$12\)</td></tr> <tr><td>\(25\)</td><td>\(\$20\)</td></tr> </table>
5511946
A snack mix uses \(3\) cups of cereal for every \(2\) cups of pretzels. One row in the table is not equivalent to this ratio. <table> <tr><th>Cereal (cups)</th><th>Pretzels (cups)</th></tr> <tr><td>\(3\)</td><td>\(2\)</td></tr> <tr><td>\(6\)</td><td>\(4\)</td></tr> <tr><td>\(9\)</td><td>\(5\)</td></tr> <tr><td>\(12\)</td><td>\(8\)</td></tr> </table> Identify the incorrect row and correct it.

Hints

- Compare each row with the original \(3:2\) relationship. - In an equivalent ratio, both quantities change by the same scale factor. - Test every row by checking whether both quantities use the same scale factor.

Solution

1. Equivalent rows must multiply both parts of \(3:2\) by the same factor. 2. The row \(6:4\) uses a factor of \(2\), and \(12:8\) uses a factor of \(4\). 3. For \(9\) cups of cereal, the factor is \(3\), so the pretzel amount must be \(2 \times 3 = 6\) cups. 4. The row \(9:5\) is incorrect; it should be \(9:6\).

Answer

The row \(9:5\) is incorrect. The corrected row is \(9:6\).
5541936
A banner design uses 3 gold stars for every 4 blue stars. Construct an equivalent-ratio table with four rows for \(3\), \(6\), \(9\), and \(12\) gold stars. Include the matching number of blue stars in each row.

Hints

- Use the given “3 for every 4” relationship as the first row. - Each new row must multiply both entries by the same scale factor. - Check that every row describes the same gold-to-blue comparison.

Solution

1. Start with the given ratio \(3:4\), so the first row is \((3,4)\). 2. Scale both quantities by \(2\) to get \((6,8)\). 3. Scale both quantities by \(3\) to get \((9,12)\). 4. Scale both quantities by \(4\) to get \((12,16)\).

Answer

<table> <tr><th>Gold stars</th><th>Blue stars</th></tr> <tr><td>3</td><td>4</td></tr> <tr><td>6</td><td>8</td></tr> <tr><td>9</td><td>12</td></tr> <tr><td>12</td><td>16</td></tr> </table>
5128116
A painter mixes blue and yellow paint to make one shade of green. The table shows mixtures that should all have the same color. <table> <tr><td>Blue paint, in fluid ounces</td><td>\(6\)</td><td>\(9\)</td><td>\(18\)</td></tr> <tr><td>Yellow paint, in fluid ounces</td><td>\(10\)</td><td>\(15\)</td><td>\(?\)</td></tr> </table> a) Show with calculations that blue paint \(\to\) yellow paint is a proportional relationship. b) Find the amount of yellow paint needed for \(18\) fluid ounces of blue paint. c) The painter wants exactly \(40\) fluid ounces of the green mixture. How many fluid ounces of blue paint should be used?

Hints

- Compare the ratio of yellow paint to blue paint in the complete columns. - Use the constant ratio for the missing table value. - Add the parts in the blue-to-yellow mixing ratio.

Solution

1. The ratios are \(10\div 6=\frac{5}{3}\) and \(15\div 9=\frac{5}{3}\), so the relationship is proportional. 2. For \(18\) fluid ounces of blue paint, the yellow amount is \(18\times \frac{5}{3}=30\) fluid ounces. 3. The blue-to-yellow ratio is \(3\) to \(5\), for \(8\) total parts. 4. Each part of a \(40\)-fluid-ounce mixture is \(40\div 8=5\) fluid ounces, so the blue amount is \(3\times 5=15\) fluid ounces.

Answer

a) The ratio of yellow to blue is constantly \(\frac{5}{3}\), so the relationship is proportional. b) \(30\) fluid ounces. c) \(15\) fluid ounces of blue paint.
5167256
Admission to a climbing park costs \(\$18\) per person. a) Make a price table for groups of \(1, 2, 3, 4, 5, 6, 7, 8, 9,\) and \(10\) people. b) A group pays \(\$126\). How many people are in the group?

Hints

- How does the price change when one more person is added? - Use your table to locate \(\$126\). - Finding the cost for \(5\) people first may help you continue the pattern.

Solution

1. Multiply each group size by \(\$18\) to complete the table. <table> <tr><td>People</td><td>\(1\)</td><td>\(2\)</td><td>\(3\)</td><td>\(4\)</td><td>\(5\)</td><td>\(6\)</td><td>\(7\)</td><td>\(8\)</td><td>\(9\)</td><td>\(10\)</td></tr> <tr><td>Price</td><td>\(\$18\)</td><td>\(\$36\)</td><td>\(\$54\)</td><td>\(\$72\)</td><td>\(\$90\)</td><td>\(\$108\)</td><td>\(\$126\)</td><td>\(\$144\)</td><td>\(\$162\)</td><td>\(\$180\)</td></tr> </table> 2. Locate \(\$126\) in the table, or calculate \(\$126 \div \$18 = 7\).

Answer

a) <table> <tr><td>People</td><td>\(1\)</td><td>\(2\)</td><td>\(3\)</td><td>\(4\)</td><td>\(5\)</td><td>\(6\)</td><td>\(7\)</td><td>\(8\)</td><td>\(9\)</td><td>\(10\)</td></tr> <tr><td>Price</td><td>\(\$18\)</td><td>\(\$36\)</td><td>\(\$54\)</td><td>\(\$72\)</td><td>\(\$90\)</td><td>\(\$108\)</td><td>\(\$126\)</td><td>\(\$144\)</td><td>\(\$162\)</td><td>\(\$180\)</td></tr> </table> b) The group has \(7\) people.
5167266
At the Mountain View Inn, breakfast costs \(\$14\) for an adult and \(\$9\) for a child. a) What is the total breakfast cost for a family with \(2\) adults and \(4\) children? b) Make a price table for \(1, 2, 3, 4,\) and \(5\) child breakfasts.

Hints

- Find the total cost for the adults. - Find the total cost for the children. - Add the two amounts. - Use multiples of \(9\) for the table.

Solution

1. Find the cost for the adults: \(2 \times \$14 = \$28\). 2. Find the cost for the children: \(4 \times \$9 = \$36\). 3. Add the costs: \(\$28 + \$36 = \$64\). 4. Complete the child-breakfast table using multiples of \(\$9\). <table> <tr><td>Child breakfasts</td><td>\(1\)</td><td>\(2\)</td><td>\(3\)</td><td>\(4\)</td><td>\(5\)</td></tr> <tr><td>Price</td><td>\(\$9\)</td><td>\(\$18\)</td><td>\(\$27\)</td><td>\(\$36\)</td><td>\(\$45\)</td></tr> </table>

Answer

a) The family’s breakfast costs \(\$64\). b) <table> <tr><td>Child breakfasts</td><td>\(1\)</td><td>\(2\)</td><td>\(3\)</td><td>\(4\)</td><td>\(5\)</td></tr> <tr><td>Price</td><td>\(\$9\)</td><td>\(\$18\)</td><td>\(\$27\)</td><td>\(\$36\)</td><td>\(\$45\)</td></tr> </table>
5190946
A hardware store sells rope for \(\$2.40\) per foot. Make a table showing the cost of \(5\,\text{ft}\), \(10\,\text{ft}\), and \(20\,\text{ft}\) of rope. A customer wants to buy \(25\,\text{ft}\) of rope and has \(\$50\). Is that enough? Justify your answer with a calculation.

Hints

- Use the price per foot to find each value in the table. - How can the cost for \(10\,\text{ft}\) help you find the cost for \(20\,\text{ft}\)? - Find the total cost of \(25\,\text{ft}\), and compare it with \(\$50\).

Solution

1. Multiply each length by \(\$2.40\) per foot to complete the table. <table> <tr><td>Rope length</td><td>\(5\,\text{ft}\)</td><td>\(10\,\text{ft}\)</td><td>\(20\,\text{ft}\)</td></tr> <tr><td>Cost</td><td>\(\$12.00\)</td><td>\(\$24.00\)</td><td>\(\$48.00\)</td></tr> </table> 2. Find the cost of \(25\,\text{ft}\): \(25 \times \$2.40 = \$60.00\). 3. Since \(\$60.00 > \$50.00\), the customer does not have enough money.

Answer

<table> <tr><td>Rope length</td><td>\(5\,\text{ft}\)</td><td>\(10\,\text{ft}\)</td><td>\(20\,\text{ft}\)</td></tr> <tr><td>Cost</td><td>\(\$12.00\)</td><td>\(\$24.00\)</td><td>\(\$48.00\)</td></tr> </table> No. The \(25\,\text{ft}\) of rope costs \(\$60.00\), which is more than \(\$50.00\).
5511956
Two tables describe batches of the same kind of drink mix, but the column order is different. <table> <tr><th colspan="2">Table A</th></tr> <tr><th>Concentrate (cups)</th><th>Water (cups)</th></tr> <tr><td>\(2\)</td><td>\(6\)</td></tr> <tr><td>\(5\)</td><td>\(15\)</td></tr> <tr><td>\(7\)</td><td>\(21\)</td></tr> </table> <table> <tr><th colspan="2">Table B</th></tr> <tr><th>Water (cups)</th><th>Concentrate (cups)</th></tr> <tr><td>\(9\)</td><td>\(3\)</td></tr> <tr><td>\(18\)</td><td>\(6\)</td></tr> <tr><td>\(24\)</td><td>\(8\)</td></tr> </table> Do the two tables represent the same concentrate-to-water relationship? Explain.

Hints

- Check the headings before comparing the ordered pairs. - Express both tables in the same concentrate-to-water order. - Simplify at least one row from each table, then check that the pattern continues.

Solution

1. In Table A, concentrate to water is \(2:6 = 1:3\). The other rows also simplify to \(1:3\). 2. Table B lists water first. Reorder a row as concentrate to water: \(3:9 = 1:3\). The other rows also give \(6:18 = 1:3\) and \(8:24 = 1:3\). 3. Both tables therefore represent the same concentrate-to-water relationship.

Answer

Yes. After accounting for the reversed column order, every row represents concentrate to water as \(1:3\).
5541946
The rows in the table must all be equivalent ratios. <table> <tr><th>Tickets</th><th>Cost</th></tr> <tr><td>6</td><td>\(\$15\)</td></tr> <tr><td>14</td><td>?</td></tr> <tr><td>?</td><td>\(\$5\)</td></tr> </table> Find both missing entries using multiplicative scale factors between rows.

Hints

- A scale factor can be greater than \(1\) or less than \(1\). - Compare a known entry with the target entry in the same column. - Whatever multiplicative factor changes one entry in a row must change the other entry by the same factor.

Solution

1. From \(6\) tickets to \(14\) tickets, the scale factor is \(\frac{14}{6}=\frac{7}{3}\). 2. Apply the same factor to the cost: \(\$15 \times \frac{7}{3}=\$35\). 3. From \(\$15\) to \(\$5\), the scale factor is \(\frac{1}{3}\). 4. Apply the same factor to the ticket count: \(6 \times \frac{1}{3}=2\).

Answer

The missing cost is \(\$35\), and the missing ticket count is \(2\).
5541956
Jordan says every row in this paintbrush price table is equivalent because the number of brushes increases by 4 each time. <table> <tr><th>Paintbrushes</th><th>Cost</th></tr> <tr><td>4</td><td>\(\$14\)</td></tr> <tr><td>8</td><td>\(\$28\)</td></tr> <tr><td>12</td><td>\(\$32\)</td></tr> </table> Is Jordan correct? Identify any incorrect row, correct it, and explain why adding the same amount in one column is not enough to make ratios equivalent.

Hints

- Check each row against the first complete row rather than only looking at differences down one column. - Ask what factor changes the number of brushes from one known row to another. - An equivalent-ratio table preserves multiplication by a common factor in both columns.

Solution

1. The first row scales by a factor of \(2\) to the second row: \(4 \times 2=8\) and \(\$14 \times 2=\$28\). 2. From \(4\) brushes to \(12\) brushes, the scale factor is \(3\). 3. The matching cost must therefore be \(\$14 \times 3=\$42\), not \(\$32\). 4. Equivalent ratios require both quantities to be multiplied by the same factor. A repeated additive change in one column does not by itself preserve the ratio.

Answer

Jordan is not correct. The row \((12,\$32)\) is incorrect and should be \((12,\$42)\). Equivalent-ratio rows preserve a multiplicative relationship between both columns.
5541966
Two drink mixes use the equivalent-ratio tables below. Mix P: <table> <tr><th>Water (cups)</th><th>Concentrate (cups)</th></tr> <tr><td>6</td><td>2</td></tr> <tr><td>12</td><td>4</td></tr> <tr><td>18</td><td>6</td></tr> </table> Mix Q: <table> <tr><th>Water (cups)</th><th>Concentrate (cups)</th></tr> <tr><td>9</td><td>2</td></tr> <tr><td>18</td><td>4</td></tr> <tr><td>27</td><td>6</td></tr> </table> Which mix has the greater concentrate-to-water ratio? Use the tables to justify your answer.

Hints

- Look for rows in the two tables that have the same concentrate amount. - Compare how much water is paired with that same amount of concentrate. - Keep the requested order as concentrate to water when you write each ratio.

Solution

1. Compare rows with the same amount of concentrate. With \(2\) cups of concentrate, Mix P uses \(6\) cups of water and Mix Q uses \(9\) cups of water. 2. The concentrate-to-water ratios are \(2:6=1:3\) for Mix P and \(2:9\) for Mix Q. 3. For the same amount of concentrate, Mix P has less water, so concentrate makes up a greater share of the mix.

Answer

Mix P. Its concentrate-to-water ratio is \(1:3\), which is greater than Mix Q’s ratio of \(2:9\).
5541976
A drink recipe keeps the same ratio of syrup to water. The table shows two equivalent mixtures. The graph shows three labeled points that could represent another mixture. <table> <tr><th>Syrup (cups)</th><th>Water (cups)</th></tr> <tr><td>\(1\)</td><td>\(2\)</td></tr> <tr><td>\(3\)</td><td>\(6\)</td></tr> </table> a) What water-to-syrup ratio does the table show? b) Read the coordinates of points A, B, and C from the graph, using syrup as the x-coordinate and water as the y-coordinate. c) Which labeled point can be added as a new row of the table while keeping the same ratio? Explain.
Figure for problem 554197

Hints

- Determine the ordered relationship shown by the two table rows. - Use the graph axes to read each labeled point as \((\text{syrup},\text{water})\). - Test each graph point against the same ordered ratio as the table.

Solution

1. The table shows \(2\) cups of water for each \(1\) cup of syrup, so the water-to-syrup ratio is \(2:1\). 2. From the graph, \(A=(4,8)\), \(B=(4,6)\), and \(C=(5,8)\). 3. Point A has water-to-syrup ratio \(8:4 = 2:1\), which matches the table. 4. Points B and C do not preserve that ratio, so only A can be added as an equivalent row.

Answer

a) \(2:1\) b) \(A=(4,8)\), \(B=(4,6)\), \(C=(5,8)\) c) Point A, because \(8:4 = 2:1\).
5541986
Every row in this partial table must represent the same red-ribbon-to-gold-ribbon ratio. <table> <tr><th>Red ribbon (ft)</th><th>Gold ribbon (ft)</th></tr> <tr><td>12</td><td>18</td></tr> <tr><td>?</td><td>7.5</td></tr> <tr><td>22</td><td>?</td></tr> </table> Find both missing entries, state the simplest red-to-gold ratio, and justify why all three rows are equivalent.

Hints

- Start with the only row in which both entries are known. - Preserve the same multiplicative relationship even when the scale factor is not a whole number. - Check each completed row against the same simplest ratio.

Solution

1. The complete row gives \(12:18\), which simplifies to \(2:3\). 2. To pair with \(7.5\) feet of gold ribbon, use the red-to-gold relationship \(2:3\): \(7.5 \times \frac{2}{3}=5\) feet of red ribbon. 3. To pair with \(22\) feet of red ribbon, multiply by the gold-to-red factor \(\frac{3}{2}\): \(22 \times \frac{3}{2}=33\) feet of gold ribbon. 4. The completed rows \(12:18\), \(5:7.5\), and \(22:33\) each reduce to the same ratio \(2:3\).

Answer

The missing entries are \(5\) ft of red ribbon and \(33\) ft of gold ribbon. The simplest red-to-gold ratio is \(2:3\), and every completed row is equivalent to \(2:3\).

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